Question

A sample of gas occurs 9.0mL at a pressure of 500mmHg. A new volume of the same sample is at 750mmHg. Use two significant figures in answer on volume.

Answer 1

`\large \boxed{\text{6.0 mL} }`

Step-by-step explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

`p_(1)V_(1) = p_(2)V_(2)`

Data:

`\begin{array}{rcrrcl}p_(1)& =& \text{500 mmHg}\qquad & V_(1) &= & \text{9.0 mL} \\p_(2)& =& \text{750 mmHg}\qquad & V_(2) &= & ?\\\end{array}`

Calculations:

`\begin{array}{rcl}500 * 9.0 & =& 750V_(2)\\4500 & = & 750V_(2)\\V_(2) & = &(4500)/(750)\\\\& = & \textbf{6.0 mL}\\\end{array}\\\text{The volume of the gas is } \large \boxed{\textbf{6.0 mL}}`